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Let $G = U \times V$ where $U, V$ are nonabelian simple groups. Then $G$ has precisely four different normal subgroups.

When I was showing the above statement, I did not use the fact that $U, V$ are nonabelian, so I want to see if someone can give me a proof that uses the information so that I can easily check if I was right.

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Did you try some examples with $U$ and $V$ abelian? E.g. $G = \mathbb{Z}_2 \times \mathbb{Z}_2$? – Martin Jan 11 '13 at 5:25
Yes. For that example, we have $1 \times 1, \mathbb{Z}/2\mathbb{Z} \times 1, 1 \times \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and I believe these are all normal subgroups. – user123454321 Jan 11 '13 at 5:26
You missed one: $\{(0,0), (1,1)\}$. // See also:… – Martin Jan 11 '13 at 5:27
Oh, is it a diagonal subgroup? Thank you. – user123454321 Jan 11 '13 at 5:36
up vote 3 down vote accepted

The proof I'm thinking of does indeed use that fact that $U$ and $V$ are nonabelian.

Let $G=U\times V$ be as you say. Let $N\leq G$ be a nontrivial normal subgroup. Suppose $x=(u,v)\in N$, with $u\neq 1$. (Or $v\neq 1$, at least one of the cases must be true.) Since $U$ is simple and nonabelian, there must exist some $g\in U$ such that $gu\neq ug$, for if not, $u\in Z(U)$, in which case $Z(U)$ is a nontrivial normal subgroup of $U$.

Let $y=(g,1)$, and consider the commutator $[y,x]=y^{-1}x^{-1}yx$. Since $N$ is normal, $[y,x]\in N$. Now define $G_1=\{(u,1):u\in U\}$. Observe that $$ [y,x]=y^{-1}x^{-1}yx=(g^{-1}u^{-1}gu,1)\in G_1 $$ and $[y,x]\neq (1,1)$ since $gu\neq ug$. Thus $[y,x]\in G_1\cap N$. But then $G_1\cap N$ is a nontrivial, normal subgroup of $G_1\simeq U$, so necessarily $G_1\cap N=G_1$. Thus $G_1\leq N$.

If there exists some $(u,v)\in N$ with $v\neq 1$, then a similar argument would show $G_2\leq N$, where $G_2=\{(1,v):v\in V\}$. From this is follows that there are precisely four normal subgroups of $G$.

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I don't see how you exclude other possibilities like $H \subseteq U \times U$ with $H = \{(x, x) : x \in U\}$ with this proof. Please let me know if I am missing anything. – user123454321 Jan 13 '13 at 6:40
@GilYoungCheong If $N$ is nontrivial and contains some $(u,v)$ with $u\neq 1$, the proof shows that $G_1\subseteq N$. One possibility then is that $N=G_1$, or $N$ properly contains $G_1$, and thus some $(u,v)\in N$ with $v\neq 1$. But then $G_2\subseteq N$. So for any $u\in U$ and any $v\in V$, $(u,1)$ and $(1,v)\in N$, but this implies $(u,1)\cdot(1,v)=(u,v)\in N$, so $N=G$. Thus the four normal subgroups are $\{\{(1,1)\}, G_1, G_2, G\}$. – Ben West Jan 13 '13 at 6:50
Thank you. I like this argument more than mine. – user123454321 Jan 13 '13 at 6:54
@GilYoungCheong Thanks, happy to help. – Ben West Jan 13 '13 at 7:00

Since the proof that I have in mind is different from above, I decided to answer the question on my own as well. The answer that I was looking for was more or less what Martin commented, which is enough to find what I assumed wrong though I take blames for not mentioning that the examples are "fine" for the answer.

Let $1 \times 1 < N \unlhd G = U \times V$.

Lemma 1. Let $N_{1} := \{(x, y) \in N : y = 1_{V}\}$ and $N_{2} := \{(x, y) \in N : x = 1_{U}\}$. Then $N_{1}, N_{2} \unlhd G$.

Proof. $N_{1} \neq \emptyset$ as it has $(1, 1)$. If $(x, 1), (x', 1) \in N_{U}$, then these elements are also in $N$, so $(x, 1)(x', 1)^{-1} = (xx'^{-1}, 1) \in N$ means the same product is in $N_{1}$, so $N_{1} \leqslant G$. Moreover, we have $(u^{-1}, v^{-1})(x, 1)(u, v) = (x^{u}, 1) \in N$ for any $(u, v) \in G$, so $N_{1} \unlhd G$. An almost identical proof gives $N_{2} \unlhd G$.

Let $\pi_{1}:U \times V \rightarrow U$ and $\pi_{2}:U \times V \rightarrow V$ denote projections.

Lemma 2. $\pi_{1}(N_{1}), \pi_{1}(N) \unlhd U, \pi_{2}(N_{2}), \pi_{2}(N) \unlhd V$.

Proof. Let $x \in \pi(N_{1})$, $(u, v) \in G$. Then $(x, 1) \in N_{1}$, so $(x, 1)^{(u,v)} = (x^{u}, 1) \in N$, so this is also in $N_{1}$, which shows $x^{u} \in \pi_{1}(N_{1})$, so $\pi_{1}(N_{1}) \unlhd U$. An almost identical proof shows $\pi_{2}(N_{2}) \unlhd V$.

Now let $x \in \pi_{1}(N)$. Then we have some $y_{0} \in V$ such that $(x, y_{0}) \in N$. Thus for any $(u, v) \in G$, we have $(x, y_{0})^{(u, v)} = (x^{u}, y_{0}^{v}) \in N$, so $x^{u} \in \pi_{1}(N)$. We conclude $\pi_{1}(N) \unlhd U$. Similarly, $\pi_{2}(N) \unlhd V$.

Lemma 3. If $\pi_{1}(N) = 1_{U}$, then $N = 1 \times 1$ or $1 \times V$. And if $\pi_{2}(N) = 1_{V}$, then $N = 1 \times 1$ or $U \times 1$.

Proof. Suppose that $\pi_{1}(N) = 1_{U}$. Then $N \unlhd 1 \times V$, so $N = 1 \times 1$ or $1 \times V$, considering $\pi_{2}(N) \unlhd V$, by Lemma 2 (if $\pi_{2}(N) > 1$, then $N \simeq \pi_{2}(N)$). Similarly for the rest of the statement.

Rest of the Argument. By Lemma 3 (and Lemma 2), we can assume $\pi_{1}(N) = U, \pi_{2}(N) = V$ (otherwise, $N \in \{1 \times 1, U \times 1, 1 \times V \}$). If $\pi_{1}(N_{1}) = 1_{U}$, then for any $(x, y), (x', y) \in N$, we have $(xx'^{-1}, 1) \in N$, so $(xx'^{-1}, 1) \in N_{1}$, so $xx'^{-1} = 1$ gives $x = x'$.

Suppose that $\pi_{1}(N_{1}) = 1_{U}$. Take any $u \in U$. Since $\pi_{1}(N) = U$, take $(u, v_{0}) \in N$. For any $a \in U$, we have $(u^{a}, v_{0}) = (u, v_{0})^{(a, 1)} \in N$. Since $(u, v_{0})$ and $(u_{a}, v_{0})$ have same element in the first component, the above argument gives $u = u^{a}$. Thus $u \in Z(U) \setminus \{1_{U}\}$, and this is nonsense since $U$ is nonabelian simple (and thus $Z(U) \lhd U$ properly gives $Z(U) = 1_{U}$.) This shows $\pi_{1}(N_{1}) = U$ and similarly we can show $\pi_{2}(N_{2}) = V$. This implies that $N = U \times V = G$.

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